We know that $\frac{3}{1+3x}=3-9x+27x^2-81x^3+...$ for $ x\in\left(-\frac{1}{3},\frac{1}{3}\right)$. Using this fact, find the function that corresponds to the following series. $ 3x-\frac{9}{2}x^2+9x^3-\frac{81}{4}x^4+...$ Choose 1 answer: Choose 1 answer: (Choice A) A $ -\arctan \left( 1-3x \right)$ (Choice B) B $\arctan \left( 1+3x \right)$ (Choice C) C $ \arctan \left( 1-3x \right)$ (Choice D) D $\ln(1-3x)$ (Choice E) E $\ln \left( 1+3x \right)$ (Choice F) F $-\ln \left( 1+3x \right)$
First, note that the derivative of $ 3x-\frac{9}{2}x^2+9x^3-\frac{81}{4}x^4+...$ is $ 3-9x+27x^2-81x^3+...=\frac{3}{1+3x}$ Antidifferentiate both sides of this equation. $\int{(3-9x+27x^2-81x^3+...)}\,dx=\int{\frac{3}{1+3x}}\,dx$ On the right-hand side, use a $~u$ -substitution $~u=1+3x~$ and $~du=3dx\,$. $3x-\frac{9}{2}x^2+9x^3-\frac{81}{4}x^4+...=\ln \left( 1+3x \right)+C$ Now let $x=0$ to see that $C=0$. Thus, $ 3x-\frac{9}{2}x^2+9x^3-\frac{81}{4}x^4+...=\ln \left( 1+3x \right)$